【leetcode】【94】Binary Tree Inorder Traversal

一、问题描述

Given a binary tree, return the inorder traversal of its nodes' values.

For example:

Given binary tree 

{1,#,2,3}

,

1
\
2
/
3

return 

[1,3,2]

.

Note: Recursive solution is trivial, could you do it iteratively?

confused what 

"{1,#,2,3}"

 means? >
read more on how binary tree is serialized on OJ.

二、问题分析

①采用递归的方式②采用栈来辅助，实现非递归的方式。

三、Java AC代码

public ArrayList<Integer> inorderTraversal(TreeNode root) {
ArrayList<Integer> res = new ArrayList<Integer>();
helper(root, res);
return res;
}
private void helper(TreeNode root, ArrayList<Integer> res)
{
if(root == null)
return;
helper(root.left,res);
res.add(root.val);
helper(root.right,res);
}

public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> list = new ArrayList<Integer>();
LinkedList<TreeNode> stack = new LinkedList<TreeNode>();
TreeNode cur = root;
while(cur!=null||!stack.isEmpty()){
while(cur!=null){
stack.push(cur);
cur = cur.left;
}
if (!stack.isEmpty()) {
list.add(stack.peek().val);
cur = stack.pop().right;
}
}
return list;
}